This question apparently wants you to get comfortable
with E = m c² . But I must say, this question is a lame
way to do it.
c = 3 x 10⁸ m/s
E = m c²
1.03 x 10⁻¹³ joule = (m) (3 x 10⁸ m/s)²
Divide each side by (3 x 10⁸ m/s)²:
Mass = (1.03 x 10⁻¹³ joule) / (9 x 10¹⁶ m²/s²)
= (1.03 / 9) x (10⁻¹³ ⁻ ¹⁶) (kg)
= 1.144 x 10⁻³⁰ kg . (choice-1)
This is roughly the mass of (1 and 1/4) electrons, so it seems
that it could never happen in nature. The question is just an
exercise in arithmetic, and not a particularly interesting one.
______________________________________
Something like this could have been much more impressive:
The Braidwood Nuclear Power Generating Station in northeastern
Ilinois USA serves Chicago and northern Illinois with electricity.
<span>The station has two pressurized water reactors, which can generate
a net total of 2,242 megawatts at full capacity, making it the largest
nuclear plant in the state.
If the Braidwood plant were able to completely convert mass
to energy, how much mass would it need to convert in order
to provide the total electrical energy that it generates in a year,
operating at full capacity ?
Energy = (2,242 x 10⁶ joule/sec) x (86,400 sec/day) x (365 da/yr)
= (2,242 x 10⁶ x 86,400 x 365) joules
= 7.0704 x 10¹⁶ joules .
How much converted mass is that ?
E = m c²
Divide each side by c² : Mass = E / c² .
c = 3 x 10⁸ m/s
Mass = (7.0704 x 10¹⁶ joules) / (9 x 10¹⁶ m²/s²)
= 0.786 kilogram ! ! !
THAT should impress us ! If I've done the arithmetic correctly,
then roughly (1 pound 11.7 ounces) of mass, if completely
converted to energy, would provide all the energy generated
by the largest nuclear power plant in Illinois, operating at max
capacity for a year !
</span>
Answer:
what's that all about
hehehwhe
Explanation:
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Answer:
Explanation:
The speed of light in these mediums shall be lower than that in vacuum thus the total time light needs to cross both the media are calculated as under
Total time = Time taken through ice + Time taken through quartz
Time taken through ice = Thickness of ice / (speed of light in ice)
Thus in the same time the it would had covered a distance of
![Distance_{vaccum}=Totaltime\times V_{vaccum}\\\\Distance_{vaccum}=10^{-2}[2.20\mu _{ice+1.50\mu _{quartz}}]](https://tex.z-dn.net/?f=Distance_%7Bvaccum%7D%3DTotaltime%5Ctimes%20V_%7Bvaccum%7D%5C%5C%5C%5CDistance_%7Bvaccum%7D%3D10%5E%7B-2%7D%5B2.20%5Cmu%20_%7Bice%2B1.50%5Cmu%20_%7Bquartz%7D%7D%5D)
we have
Applying values we have
![Distance_{vaccum}=10^{-2}[2.20\times 1.309+1.50\times 1.542]](https://tex.z-dn.net/?f=Distance_%7Bvaccum%7D%3D10%5E%7B-2%7D%5B2.20%5Ctimes%201.309%2B1.50%5Ctimes%201.542%5D)
It’s C
solar
correct me if i’m wrong though
